Suppose the real valued function $f$ is differentiable on $[0,\,1]$ with bounded derivative. How do we prove the following? $$\int_0^1 f^2-\bigg(\int_0^1 f\bigg)^2\le \frac1{12}\sup_{x\in[0,1]}|f'(x)|^2.$$ I tried substituting $f(x)^2=f(0)^2+2\int_0^x f(t)f'(t)\,dt$, but it did not seem to work.


2 Answers 2


We have \begin{align*} \left(f(x)-\int_{0}^{1}f(t)dt\right)^{2}&=\left(\int_{0}^{1}(f(x)-f(t))dt\right)^{2}\\ &=\left(\int_{0}^{1}(x-t)f'(\xi_{x,t})dt\right)^{2}\\ &\leq\|(f')^{2}\|_{\infty}\left(\int_{0}^{1}(x-t)dt\right)^{2}, \end{align*} now \begin{align*} \int_{0}^{1}\left(\int_{0}^{1}(x-t)dt\right)^{2}dx=\dfrac{1}{12}, \end{align*} and note that $\text{Var}(X)=E(X-E(X))^{2}$.

Here I have used a version of Mean Value Theorem: \begin{align*} \int_{a}^{b}u(x)v(x)dx=c\int_{a}^{b}u(x)dx, \end{align*} where $m:=\inf v\leq c\leq M:=\sup v$, so in this context, \begin{align*} \int_{0}^{1}(x-t)f'(\xi_{x,t})dt=c\int_{0}^{1}(x-t)dt, \end{align*} where $-\sup|f'|=\inf(-|f'|)\leq\inf f'\leq c\leq\sup f'\leq\sup|f'|$, so $c^{2}\leq(\sup|f'|)^{2}=\sup|f'|^{2}:=\|(f')^{2}\|_{\infty}$.


Let $X \sim U(0,1)$, then we may write the left-hand side as $\mathbb{V}(f(X))$. Now the variance is invariant under translations, so $\mathbb{V}(f(X)) = \mathbb{V}(f(X)-f(1/2))$. This yields \begin{align}\mathbb{V}(f(X)-f(1/2)) &= \int_{0}^{1}(f(x)-f(1/2))^2 dx - \left(\int_{0}^{1} f(x) - f(1/2)dx\right)^2 \\ &\leq \int_{0}^{1} (f'(\xi_x)(x-1/2))^2 dx \leq \sup_{x\in [0,1]}|f'(x)|^2\int_{0}^{1}(x-1/2)^2 dx \\ &= \frac{1}{12}\sup_{x\in [0,1]}|f'(x)|^2 \\ & \end{align}

  • $\begingroup$ +1 Very slick! If I could accept both yours and user284331's answer, I would. I accept his since it came in a hair earlier. $\endgroup$
    – Hans
    Dec 22, 2017 at 1:05

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